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Solution for If the Vectors ^ I − 2 X ^ J + 3 Y ^ K and ^ I + 2 X ^ J − 3 Y ^ K Are Perpendicular, Then the Locus of (X, Y) is - CBSE (Commerce) Class 12 - Mathematics

Question

If the vectors $\hat{i} - 2x \hat{j} + 3y \hat{k} \text{ and } \hat{i} + 2x \hat{j} - 3y \hat{k}$ are perpendicular, then the locus of (xy) is

• (a) a circle

• (b) an ellipse

• (c) a hyperbola

• (d) None of these

Solution

(b) an ellipse

$\text{ Let }, \vec{a} = \hat{ i } - 2x \hat{j} + 3y \hat{k} \text{ and } \vec{b} = \hat{i} + 2x \hat{j} - 3y \hat{k}$
$\text{ It is given that the vectors are perpendicular. So, their dot product is zero }.$
$\vec{a} . \vec{b} = 0$
$\Rightarrow \left( \hat{i} - 2x \hat{j} + 3y \hat{k} \right) . \left( \hat{i} + 2x \hat{j} - 3y \hat{k} \right) = 0$
$\Rightarrow 1 - 4 x^2 - 9 y^2 = 0$
$\Rightarrow 4 x^2 + 9 y^2 = 1$
$\text{ Dividing both sides by } 36, \text{ we get }$
$\frac{x^2}{9} + \frac{y^2}{4} = 1$
$\text{ This is an ellipse }.$

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Solution If the Vectors ^ I − 2 X ^ J + 3 Y ^ K and ^ I + 2 X ^ J − 3 Y ^ K Are Perpendicular, Then the Locus of (X, Y) is Concept: Introduction of Vector.
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